Question

In an acidic medium, 10 mL of 0.25 M oxalic acid is titrated with \(KMnO_4\) solution. If the volume of \(KMnO_4\) solution required to reach the end point is 10 mL, the strength of the \(KMnO_4\) solution is

• A. \(0.15\,M\)
• B. \(0.10\,M\)
• C. \(0.20\,M\)
• D. \(0.25\,M\)

Hint:

For acidic \(KMnO_4\), \[ n\text{-factor}=5 \] For oxalic acid, \[ n\text{-factor}=2 \] Always convert molarity to normality before applying \(N_1V_1=N_2V_2\).

Answer 1

The Correct Option is B

Step 1: Set up the equivalence relation.
At the end point of a redox titration, equivalents of oxidant equal equivalents of reductant: \[ N_1 V_1 = N_2 V_2 \]
Step 2: Find the n-factor of oxalic acid.
Oxalic acid \((H_2C_2O_4)\) is the reducing agent and loses 2 electrons, so its n-factor is 2.
Step 3: Get the normality of oxalic acid.
\[ N_1 = M \times n = 0.25 \times 2 = 0.50\,N \]
Step 4: Solve for the normality of \(KMnO_4\).
With \(V_1 = V_2 = 10\) mL: \[ 0.50 \times 10 = N_2 \times 10 \implies N_2 = 0.50\,N \]
Step 5: Find the n-factor of \(KMnO_4\).
In acid medium \(MnO_4^-\) goes from +7 to +2, a gain of 5 electrons, so its n-factor is 5.
Step 6: Convert normality to molarity.
\[ M = \frac{N}{n} = \frac{0.50}{5} = 0.10\,M \]
\[ \boxed{0.10\,M} \]